Give an $$ \varepsilon-\delta $$ proof of the limit fact. $$ \lim _{x \rightarrow 0}(6 x+7)=7 $$ Let $$ \varepsilon0 $$ be given. Choose the correct answer below. A. Choose $$ \delta=\frac{\varepsilon}{6} $$. Then for all $$ x $$ satisfying $$ 0

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Give an $$ \varepsilon-\delta $$ proof of the limit fact. $$ \lim _{x \rightarrow 0}(6 x+7)=7 $$ Let $$ \varepsilon>0 $$ be given. Choose the correct answer below. A. Choose $$ \delta=\frac{\varepsilon}{6} $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(6 x+7)-7|=|6 x|=6|x|<6 \delta=\varepsilon $$ holds. B. Choose $$ \delta=\varepsilon $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(6 x+7)-7|=|6 x|<\delta=\varepsilon $$ holds. C. Choose $$ \delta=\frac{\varepsilon}{7} $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(6 x+7)-6 x|=|7 x|=7|x|<7 \delta=\varepsilon $$ holds. D. Choose $$ \delta=6 \varepsilon $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(6 x+7)-7|=|6 x|=6|x|<\frac{\delta}{6}=\varepsilon $$ holds.

UpStudy AI · Accepted Answer

We need to show that for every ε > 0, there exists a δ > 0 such that whenever 0 < |x – 0| < δ, it follows that |(6x + 7) – 7| < ε. Notice that |(6x + 7) – 7| = |6x| = 6|x|. To make 6|x| < ε, we can require that |x| < ε/6. Therefore, if we choose δ = ε/6, then for all x satisfying 0 < |x| < δ we have: 6|x| < 6(δ) = 6(ε/6) = ε. Thus, the proof is complete using the choice δ = ε/6, which corresponds to option A. Choose $$ \delta = \frac{\varepsilon}{6} $$. Then for all $$ x $$ satisfying $$ 0 < |x| < \delta $$, the inequality $$ |6x| < \varepsilon $$ holds.

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